\(\dfrac{3}{8}\times\dfrac{5}{12}+\dfrac{5}{8}\times\dfrac{5}{12}\)

\(=\dfrac{5}{12}\times\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\)

\(=\dfrac{5}{12}\times1\)

\(=\dfrac{5}{12}\)

3/8 x 5/12 + 5/8 x 5/12 

= 5/12 x (3/8 + 5/8 )

= 5/12 x 1 

= 5/12 

\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{15}+....+\dfrac{1}{45}\)

Đề bài như này thì ko tính đc nhé e phải như này mới tính đc viết nhầm đề rồi hay sao ý 

\(A=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+....+\dfrac{1}{45}\)

\(A=2\times(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+....+\dfrac{1}{90})\\\)

\(A=2\times\left(\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+...+\dfrac{1}{9\times10}\right)\)

\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+.....+\dfrac{1}{9}-\dfrac{1}{10}\right)\)

\(A=2\times\left(\dfrac{1}{2}-\dfrac{1}{10}\right)\)

\(A=2\times\dfrac{2}{5}\)

\(A=\dfrac{4}{5}\)

\(\dfrac{6}{3\cdot5}+\dfrac{6}{5\cdot7}+\dfrac{6}{7\cdot9}+.....+\dfrac{6}{33\cdot35}\)

\(=\left(\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{33\cdot35}\right)\cdot3\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+.....+\dfrac{1}{33}-\dfrac{1}{35}\right)\cdot3\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{35}\right)\cdot3\)

\(=\dfrac{32}{3\cdot35}\cdot3\)

\(=\dfrac{32}{35}\)

\(A=\dfrac{202\cdot209+10}{202\cdot211-394}\)

\(A=\dfrac{202\cdot209+10}{202\cdot209+2\cdot202-394}\)

\(A=\dfrac{202\cdot209+10}{202\cdot209+10}\\ A=1\)

\(x^2-xy+x-y\)

\(=x\left(x-y\right)+\left(x-y\right)\)

\(=\left(x+1\right)\left(x-y\right)\)

\(x^2-xy+x-y\)

\(=x\left(x-y\right)+\left(x-y\right)\)

\(=\left(x+1\right)\left(x-y\right)\)

\(x-\dfrac{5}{15}=\dfrac{42}{105}\)

\(\Rightarrow x=\dfrac{42}{105}+\dfrac{5}{15}\)

\(\Rightarrow x=\dfrac{11}{15}\)

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